Assignment 22 Q2iv

Submitted by mrmonkey on Tue, 10/17/2017 - 16:01

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I don't understand how you can show that $e^xe^{-x} = 1$ ? I understand that because of the differentiation it must be a constant, and that when $x=0 , e^x=1$ , but how does this show that $e^xe^{-x} = 1$ ? I am possibly just being thick, or perhaps something is assumed which I thought wasn't meant to be?

Constant function

Permalink Submitted by cg213 on Wed, 10/18/2017 - 11:36

We have that the derivative is constant (for anyone stuck here, take $a=1$ and $b=-1$ in the first result in this part).

Hence we have $e^xe^{-x}=c$ for some $c$ . Substituting $x=0$ gives:
$\begin{align*} e^0e^{-0}&=c\\ e^0e^0&=c\\ 1 \times 1 &=c\\ 1&=c \end{align*}$
Hence we have $c=1$ and $e^xe^{-x}=1$ .